Thanks for everyone.
I think the approach below is most suitable for me, as a beginner.
x=list()
> > for(i in 1:n){
> > x[[i]]=rnorm(i,0,1)
> > }
Now, I am trying to obtain the sample variance (S^2) of the 1000 samples that I
have generated before.
I am wondering what command I should use in order to get the sample variance
for all the 1000 samples.
What I am capable of doing now is just typing in
var(z[[1]])
var(z[[2]]).....................
Thanks for help.
Debbie
> Date: Thu, 14 May 2009 08:26:03 +0200
> From: waclaw.marcin.kusnierc...@idi.ntnu.no
> To: b.rowling...@lancaster.ac.uk
> CC: r-help@r-project.org
> Subject: Re: [R] Simulation
>
> Barry Rowlingson wrote:
> > On Wed, May 13, 2009 at 9:56 PM, Wacek Kusnierczyk
> > <waclaw.marcin.kusnierc...@idi.ntnu.no> wrote:
> >
> >> Barry Rowlingson wrote:
> >>
> >
> >
> >> n = 1000
> >> benchmark(columns=c('test', 'elapsed'), order=NULL,
> >> 'for'={ l = list(); for (i in 1:n) l[[i]] = rnorm(i, 0, 1) },
> >> lapply=lapply(1:n, rnorm, 0, 1) )
> >> # test elapsed
> >> # 1 for 9.855
> >> # 2 lapply 8.923
> >>
> >>
> >>
> >>> Yes, you can probably vectorize this with lapply or something, but I
> >>> prefer clarity over concision when dealing with beginners...
> >>>
> >> but where's the preferred clarity in the for loop solution?
> >>
> >
> > Seriously? You think:
> >
> > lapply(1:n, rnorm, 0, 1)
> >
> > is 'clearer' than:
> >
> > x=list()
> > for(i in 1:n){
> > x[[i]]=rnorm(i,0,1)
> > }
> >
> > for beginners?
> >
>
> seriously, i do; but it does depend on who those beginners are. if
> they come directly from c and the like, you're probably right.
>
>
> > Firstly, using 'lapply' introduces a function (lapply) that doesn't
> > have an intuitive name. Also, it takes a function as an argument. The
> > concept of having a function as a parameter to another function is
> > something that a lot of programming beginners have trouble with -
> > unless they were brought up on LISP of course, and few of us are.
> >
>
> well, that's one of the first things you learn on a programming
> languages course that is not procedural programming-{centered,biased}.
> no need for prior lisp experience. if messing with closures in not
> involved (as here), no need for advanced discussion is needed.
>
> also, the for looping may not be as trivial stuff to explain as you
> might think. note, you're talking about r, not c, and the treatment of
> iterator variables in for loops in scripting languages differs:
>
> perl -e '
> $i = 0;
> for $i (1..5) { # iterate with $i
> };
> print "$i\n" '
> # 0
>
> ruby -e '
> i = 0
> for i in 1..5 # iterate with i
> end
> printf "%d\n", i '
> # 5
>
> and you've gotten into explaining lexical scoping etc.
>
>
> > I propose that the for-loop example is clearer to a larger population
> > than the lapply version.
>
> which population have you sampled from? you may not be wrong, but give
> some data.
>
>
> > Plus it's only useful in that form if the
> > first parameter is the one you want to lapply over. If you want to
> > work over the third parameter, say, you then need:
> >
> > lapply(1:n,function(i){rnorm(100,0,i)})
> >
> > at which point you've introduced anonymous functions. The jump from:
> >
> > x[[i]] = rnorm(i,0,1)
> > to
> > x[[i]] = rnorm(100,0,i)
> >
> > is much less than the changes in the lapply version, where you have to
> > go 'oh hang on, lapply only works on the first argument, so you have
> > to write another function, but you can do that inline like this...'.
> >
>
> you may be unhappy to learn that you're unaware of how the lapply
> solution can still be nicely adapted here:
>
> lapply(1:n, rnorm, n=100, mean=0)
>
> > Okay, maybe my example is a little contrived, but I still think for a
> > beginners context it's important not to jump too many paradigms at a
> > time.
> >
>
> for a complete beginner, jump into for loops may not be that trivial as
> you seem to think. there's still quite some stuff to be explained to
> clarify that
>
> i = 0
> for (i in 1:n)
> # do stuff
> print(i)
>
> will print n, not 0. unless n=0, of course.
>
> vQ
>
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